Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
LESSON 2-5: REASONING IN ALGEBRA AND GEOMETRY Goal: To connect reasoning in Algebra and Geometry. PROPERTIES OF EQUALITY: Let π, π, and π be any real numbers. Addition Property Subtraction Property Multiplication Property Division Property Reflexive Property Symmetric Property Transitive Property Substitution Property πΌπ π = π, π‘βππ π + π = π + π. πΌπ π = π, π‘βππ π β π = π β π. πΌπ π = π, π‘βππ π β π = π β π. πΌπ π = π πππ π β 0, π‘βππ π = π. π=π πΌπ π = π, π‘βππ π = π. πΌπ π = π πππ π = π, π‘βππ π = π. πΌπ π = π, π‘βππ π πππ πππππππ π ππ πππ¦ ππ₯ππππ π πππ. Distributive Property π π + π = ππ + ππ π π π π β π = ππ β ππ EXAMPLE 1: JUSTIFYING STEPS WHEN SOLVING AN EQUATION ο’ What is the Value of x? Justify each step. β π΄ππ πππ β πππΆ are supplementary β βs that form a linear pair are supplementary Definition of supplementary β β² π πβ π΄ππ + πβ πππΆ = 180 Substitution property 2π₯ + 30 + π₯ = 180 Combining like terms 3π₯ + 30 = 180 Subtraction Property of Equality 3π₯ = 150 Division Property of Equality π₯ = 50 PROPERTIES OF CONGRUENCE: Reflexive Property π΄π΅ β π΄π΅ Symmetric Property πΌπ π΄π΅ β πΆπ·, π‘βππ πΆπ· β π΄π΅ πΌπ β π΄ β β π΅, π‘βππ β π΅ β β π΄ Transitive Property πΌπ π΄π΅ β πΆπ· πππ πΆπ· β πΈπΉ, π‘βππ π΄π΅ β πΈπΉ πΌπ β π΄ β β π΅ πππ β π΅ β β πΆ, π‘βππ β π΄ β β πΆ. β π΄ β β π΄ EXAMPLE 2: USING PROPERTIES OF EQUALITY AND CONGRUENCE ο’ What is the property of congruence that justifies going from the first statement to the second statement? ο 2π₯ + 9 = 19 2π₯ = 10 ο β π β β π πππ β π β β πΏ β π β β πΏ Transitive Prop. Of Congruence ο πβ πΈ = πβ π πβ π = πβ πΈ Symmetric Prop. Of Equality Subtraction Prop. Of Equality EXAMPLE 3: WRITING A TWO-COLUMN PROOF ASSIGNMENT: ο’ Pg 117 β 118 #βs 6-12 even, 14-19 all and 25